\(\int e^{a+b x} \sinh (a+b x) \, dx\) [304]

Optimal result

Integrand size = 14, antiderivative size = 23 \[ \int e^{a+b x} \sinh (a+b x) \, dx=\frac {e^{2 a+2 b x}}{4 b}-\frac {x}{2} \]

[Out]

1/4*exp(2*b*x+2*a)/b-1/2*x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2320, 12, 14} \[ \int e^{a+b x} \sinh (a+b x) \, dx=\frac {e^{2 a+2 b x}}{4 b}-\frac {x}{2} \]

[In]

Int[E^(a + b*x)*Sinh[a + b*x],x]

[Out]

E^(2*a + 2*b*x)/(4*b) - x/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {-1+x^2}{2 x} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {\text {Subst}\left (\int \frac {-1+x^2}{x} \, dx,x,e^{a+b x}\right )}{2 b} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {1}{x}+x\right ) \, dx,x,e^{a+b x}\right )}{2 b} \\ & = \frac {e^{2 a+2 b x}}{4 b}-\frac {x}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int e^{a+b x} \sinh (a+b x) \, dx=\frac {e^{2 a+2 b x}}{4 b}-\frac {x}{2} \]

[In]

Integrate[E^(a + b*x)*Sinh[a + b*x],x]

[Out]

E^(2*a + 2*b*x)/(4*b) - x/2

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83

method	result	size
risch	\(\frac {{\mathrm e}^{2 b x +2 a}}{4 b}-\frac {x}{2}\)	\(19\)
derivativedivides	\(\frac {\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}-\frac {b x}{2}-\frac {a}{2}+\frac {\cosh \left (b x +a \right )^{2}}{2}}{b}\)	\(37\)
default	\(\frac {\frac {\cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{2}-\frac {b x}{2}-\frac {a}{2}+\frac {\cosh \left (b x +a \right )^{2}}{2}}{b}\)	\(37\)

[In]

int(exp(b*x+a)*sinh(b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/4*exp(2*b*x+2*a)/b-1/2*x

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (18) = 36\).

Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.17 \[ \int e^{a+b x} \sinh (a+b x) \, dx=-\frac {{\left (2 \, b x - 1\right )} \cosh \left (b x + a\right ) - {\left (2 \, b x + 1\right )} \sinh \left (b x + a\right )}{4 \, {\left (b \cosh \left (b x + a\right ) - b \sinh \left (b x + a\right )\right )}} \]

[In]

integrate(exp(b*x+a)*sinh(b*x+a),x, algorithm="fricas")

[Out]

-1/4*((2*b*x - 1)*cosh(b*x + a) - (2*b*x + 1)*sinh(b*x + a))/(b*cosh(b*x + a) - b*sinh(b*x + a))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (15) = 30\).

Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.74 \[ \int e^{a+b x} \sinh (a+b x) \, dx=\begin {cases} \frac {x e^{a} e^{b x} \sinh {\left (a + b x \right )}}{2} - \frac {x e^{a} e^{b x} \cosh {\left (a + b x \right )}}{2} + \frac {e^{a} e^{b x} \cosh {\left (a + b x \right )}}{2 b} & \text {for}\: b \neq 0 \\x e^{a} \sinh {\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(exp(b*x+a)*sinh(b*x+a),x)

[Out]

Piecewise((x*exp(a)*exp(b*x)*sinh(a + b*x)/2 - x*exp(a)*exp(b*x)*cosh(a + b*x)/2 + exp(a)*exp(b*x)*cosh(a + b*
x)/(2*b), Ne(b, 0)), (x*exp(a)*sinh(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int e^{a+b x} \sinh (a+b x) \, dx=-\frac {1}{2} \, x - \frac {a}{2 \, b} + \frac {e^{\left (2 \, b x + 2 \, a\right )}}{4 \, b} \]

[In]

integrate(exp(b*x+a)*sinh(b*x+a),x, algorithm="maxima")

[Out]

-1/2*x - 1/2*a/b + 1/4*e^(2*b*x + 2*a)/b

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int e^{a+b x} \sinh (a+b x) \, dx=-\frac {2 \, b x + 2 \, a - e^{\left (2 \, b x + 2 \, a\right )}}{4 \, b} \]

[In]

integrate(exp(b*x+a)*sinh(b*x+a),x, algorithm="giac")

[Out]

-1/4*(2*b*x + 2*a - e^(2*b*x + 2*a))/b

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int e^{a+b x} \sinh (a+b x) \, dx=\frac {{\mathrm {e}}^{2\,a+2\,b\,x}}{4\,b}-\frac {x}{2} \]

[In]

int(exp(a + b*x)*sinh(a + b*x),x)

[Out]

exp(2*a + 2*b*x)/(4*b) - x/2